How to : $\int_1^3 \sqrt{x^4+1}dx$ 
Calculate the integral :
$$\int_1^3 \sqrt{x^4+1}dx$$

I need to solve this integral. I've tried to do a trigonometric substitution but it didn't work out for me, and also tried to use a integral calculator from the internet but it couldn't solve it.
Thank you in advance.
 A: This integral is not an elementary one (meaning it cannot be described as a solution in terms of standard mathematical functions), as its solution involves elliptic functions.
The result is :
$$\int_1^3 \sqrt{x^4+1}dx=\frac{1}{3}\sqrt2[-1 + 3\sqrt{41} + (1+i)F(i\sinh^{-1}(\sqrt[4]{-1})|-1) - (1+i)F(i\sinh^{-1}\sqrt[4]{-1})|-1)] \approx 8.98006$$
where $F(x|m)$ is the elliptic integral of the first kind with parameter $m=k^2$.
If that seems like nonsense to you, double check your calculations that lead to this integral. 
If not, you'll want to use a numerical method to calculate this, like the trapezoid rule, as mentioned in the comments as well, which states : 

$$\int_a^bf(x)dx \approx(b-a)\frac{f(a)+f(b)}{2}$$

Applying that, yields : 
$$\int_1^3 \sqrt{x^4+1}dx \approx(3-1)\frac{f(1)+f(3)}{2}=f(1)+f(3)= \sqrt{2}+\sqrt{3^4+1} \approx10.46$$
Note the difference between the $2$ calculations, that's a relative error right there. There are also more methods, that give you better approximation. You can check about these around the internet or in a book that studies Numerical Analysis .
Update :
Simpson's Rule (as mentioned in your comment) states that : 

$$\int_a^bf(x)dx \approx \frac{b-a}{6}\bigg[f(a) + 4f\bigg(\frac{a+b}{2}\bigg) + f(b)\bigg]$$

Can you use that and derive your result ?
